Question: $f(n) = -n$ $h(n) = -5n-3(f(n))$ $g(x) = 2x^{2}+6x-4(f(x))$ $ h(g(-5)) = {?} $
Solution: First, let's solve for the value of the inner function, $g(-5)$ . Then we'll know what to plug into the outer function. $g(-5) = 2(-5)^{2}+(6)(-5)-4(f(-5))$ To solve for the value of $g$ , we need to solve for the value of $f(-5)$ $f(-5) = -(-5)$ $f(-5) = 5$ That means $g(-5) = 2(-5)^{2}+(6)(-5)+(-4)(5)$ $g(-5) = 0$ Now we know that $g(-5) = 0$ . Let's solve for $h(g(-5))$ , which is $h(0)$ $h(0) = (-5)(0)-3(f(0))$ To solve for the value of $h$ , we need to solve for the value of $f(0)$ $f(0) = -0$ $f(0) = 0$ That means $h(0) = (-5)(0)+(-3)(0)$ $h(0) = 0$